# Pentagon Problem

• Jun 19th 2012, 11:28 AM
Telo
Pentagon Problem
Attachment 24112

Pentagon ABCDE

BC = CD = BD = AE

Perimeter (ABCDE) = 10

Find the sides of rectangle, for which the area of the pentagon will be maximum.

(sry for my english)
• Jun 19th 2012, 12:08 PM
richard1234
Re: Pentagon Problem
Let $BC = x$ and $AB = y$. The area of triangle BCD is $\frac{x^2 \sqrt{3}}{4}$, and the area of ABDE is $xy$. Hence the area of the pentagon P is

$[P] = \frac{x^2 \sqrt{3}}{4} + xy = \frac{x^2 \sqrt{3} + 4xy}{4}$

However you know that the perimeter is 10, so $3x + 2y = 10 \Rightarrow y = \frac{10-3x}{2}$. Substitute into the area equation to obtain

$[P] = \frac{x^2 \sqrt{3} + 4x(\frac{10-3x}{2}))}{4}$

Simplify, and differentiate both sides with respect to x and find critical points.
• Jun 19th 2012, 12:11 PM
Telo
Re: Pentagon Problem
the answer is 10/(6-sqr(3)) and (15-5sqr(3))/(6-sqr(3)) but i cant get those answers :(
• Jun 19th 2012, 12:21 PM
Reckoner
Re: Pentagon Problem
Quote:

Originally Posted by Telo
Find the sides of rectangle, for which the area of the pentagon will be maximum.

Suppose that the vertical sides of the rectangle each have length $a,$ and suppose that the other sides have length $b.$

The area of the pentagon is

$A = ab + \frac12b\left(\frac{\sqrt3}2b \right)$

$\Rightarrow A = ab + \frac{b^2\sqrt3}4,$

and the perimeter is

$P = 2a + 3b = 10$

$\Rightarrow b = \frac13\left(10 - 2a\right).$

Substituting this into the area equation above produces

$A = a\left[\frac13(10-2a)\right] + \left[\frac13(10-2a)\right]^2\frac{\sqrt3}4$

$= \frac19\left[\left(\sqrt3-6\right)a^2 + \left(30 - 10\sqrt3\right)a + 25\sqrt3\right]$

Differentiating,

$\frac{dA}{da} = \frac19\left[\left(2\sqrt3 - 12\right)a + 30 - 10\sqrt3\right]$

We locate the critical value:

$\frac{dA}{da} = 0$

$\Rightarrow\frac19\left[\left(2\sqrt3 - 12\right)a + 30 - 10\sqrt3\right]=0$

$\Rightarrow\left(2\sqrt3 - 12\right)a = 10\sqrt3 - 30$

$\Rightarrow a = \frac{10\sqrt3 - 30}{2\sqrt3 - 12} = \frac{5\sqrt3 - 15}{\sqrt3 - 6}$

$\Rightarrow a = \frac{15 - 5\sqrt3}{6 - \sqrt3}$

Now you can find $b.$
• Jun 19th 2012, 12:28 PM
Reckoner
Re: Pentagon Problem
Quote:

Originally Posted by richard1234
However you know that the perimeter is 10, so $4x + 2y = 10 \Rightarrow y = 5-2x$.

Small correction: that should be $3x + 2y = 10.$ We want the perimeter of the pentagon, so we just need the outer sides.
• Jun 19th 2012, 12:34 PM
Telo
Re: Pentagon Problem
thanks alot!
• Jun 19th 2012, 01:24 PM
richard1234
Re: Pentagon Problem
@Reckoner whoops. Can't do math in my head anymore lol. I just fixed my original post.